Question

A 20.0 mL sample of 0.200 molar K2CO3 solution is added to 30.0 mL of 0.400 molar Ba(NO3)2
solution. Barium carbonate precipitates. The concentration of Barium ion, Ba+2, in the solution,
after reaction is

Answer 1

Answer:

The reaction for potassium carbonate and barium nitrate to form barium carbonate is

Ba(NO3)2(aq)+K2CO3(aq)→BaCO3(s)+2KNO3Ba(NO3)2(aq)+K2CO3(aq)→BaCO3(s)+2KNO3

In the compound barium carbonate there is a 1 : 1 ratio of barium ions to carbonate ions.

To determine which reactant is in excess and which is limiting it is necessary to calculate the moles of carbonate ions and barium ions.

0.020L×0.2molCO2−3L=0.004molCO2−30.030L×0.4Ba2+=0.012molBa2+0.020L×0.2molCO32−L=0.004molCO32−0.030L×0.4Ba2+=0.012molBa2+

The carbonate is the limiting reactant. The amount of barium that remains in solution after the reaction is 0.012 moles - 0.004 moles = 0.080 mole barium.

The concentration of barium left in solution is the moles of barium divided by the volume of solution.

0.008mol0.02L+0.03L=0.160M

Explanation:

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